# Oscillating Cylinder in Parallel Flow: Analytical Solution

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 July 7, 2010, 00:17 Oscillating Cylinder in Parallel Flow: Analytical Solution #1 New Member   Join Date: Jul 2010 Posts: 8 Rep Power: 8 I need a paper (lecture or class notes) showing a basic analytical solution to parallel flow past an oscillating cylinder. The oscillation can be transverse or parallel to the flow. The flow is laminar, incompressible, inviscid, and unsteady. So basically, I want to solve Laplace's equation for a moving cylinder. I believe the unsteady nature of the flow appears in the momentum equation, which for this problem reduces to an unsteady Bernoulli's equation. Ostensibly, the potential flow solution around a stationary cylinder is quite simple and easily accessible as contained in any introductory fluids text. But I've had no luck whatsoever finding the analytical solution for the moving cylinder. I have found a bunch of papers with a numerical solution though. Feel free to send those along as well. Related question just to make sure: For the problem stated (inviscid, incompressible), I won't be solving for vortex shedding? Need to include the viscous term along with the convection term to do so? Thanks for the help. [I need it for background of my research.]

 July 7, 2010, 08:30 #2 Senior Member   Join Date: Jul 2009 Posts: 232 Rep Power: 11 According to theory, there is no vortex shedding if the flow is inviscid. The flow all separates at the rear stagnation point. Conversely, there is no general theoretical solution for viscous flow over a cylinder where separation and shedding occur.

 July 7, 2010, 09:33 #3 Super Moderator     Praveen. C Join Date: Mar 2009 Location: Bangalore Posts: 255 Blog Entries: 6 Rep Power: 10 If you assume potential flow, at any instant, you have Laplace equation for the potential with some Neumann conditions on the cylinder and Dirichlet conditions in the farfield. The potential adjusts instantaneously to the boundary motion. To find the solution, you can transform to a frame in which the cylinder is at rest at that instant, use the standard solution for steady cylinder, and then transform back to your original frame. Once you have the potential, you get the pressure from the Bernoulli equation. If you get the solution, please write about it here.

July 7, 2010, 17:05
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 The potential adjusts instantaneously to the boundary motion. Once you have the potential, you get the pressure from the Bernoulli equation.
I'm not sure about the above. Unsteady Bernoulli's eqn is of the form:

dphi/dt + p/pho + 1/2V^2 + gz = constant

So you can't just solve for the potential at each time step because of the term dphi/dt. The change in the potential from the previous time step matters for the solution.

I've looked more and I still can't find an analytical solution for this problem. Anyone else have luck?

Also, how exactly would I approach this numerically? If I'm using a Galerkin/FEM approach, should my test functions be a seperation of variables type function, i.e. X(x)Y(y)T(t)? The continuity equation reduces to Laplace's equation which is only a function of the 2D spatial variables. But then the unsteady Bernoulli's equation (reduction of the momentum equation) is a function of both space and time.

I've only previously solved steady heat conduction and stress strain problems that included just one equation to satisfy. A little unsure about taking it to 2 equations and temporal discretization in an FEM context (I've done finite difference but that was rather trivial).

 July 8, 2010, 00:32 #5 Super Moderator     Praveen. C Join Date: Mar 2009 Location: Bangalore Posts: 255 Blog Entries: 6 Rep Power: 10 The potential is determined from Laplace equation and time enters into the picture through the boundary conditions. So when you get your potential it is . Then this can be plugged into the Bernoulli equation to get the pressure.

July 8, 2010, 12:14
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 Originally Posted by praveen The potential is determined from Laplace equation and time enters into the picture through the boundary conditions. So when you get your potential it is . Then this can be plugged into the Bernoulli equation to get the pressure.
But by solving it in this manner, you're obscuring the unsteady nature of the system. Let's say its heat. The heat conduction starts off at some initial distribution then "flows" through the system during the transient period.

This unsteady "flow" is governed by the derivative with respect to time. By doing it your way (solving the Laplace equation for the instantaneous steady state according to the current BC's), you can't know the dphi/dt term. You've ignored everything in the past and you can't do that when a time derivative is present.

The two equations (Laplace and unsteady Bernoulli) must be solved simultaneously.

 July 8, 2010, 14:50 #7 Senior Member   Join Date: Jul 2009 Posts: 232 Rep Power: 11 Praveen is correct - the Laplace equation can be solved for the velocity potential, and the only time dependence enters in through the boundary conditions. Bernoulli's equation is then used to compute the pressure once you have the potential. The first thing I would try is to simply separate the variables - assume phi(x,t) = f(x)g(t). Then clearly Lap(f) = 0, and if the boundary conditions allow separation, you should be able to solve this equation for f, with g(t) being determined by the time-dependent part of the BC. Then use Bernoulli's equation to find the pressure field p(x,t) for any value of time. Note that this is not a formal solution procedure, but simply a suggestion of one possible approach. Also note that this does not ignore the fact that phi depends on time, but makes use of the fact that the velocity potential can only have a certain form of dependence on time based on the fact that it satisfies Lap(phi) = 0 (no time derivative in this equation). Last edited by agd; July 8, 2010 at 15:05.